Question

An emf of 17.0 mV is induced in a 483-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic flux through each turn of the coil at an instant when the current is 4.30 A?

Answer 1

Answer:

$0.0000151Tm^{2}$

Explanation:

$E=17.0mV=0.017V\\t=10s\\I=4.30A$

$N=483 turns$

The relation to find the magnetic flux is:

φ$=\frac{LI}{N}$

so we need to find L

$L=\frac{E}{dI/dt}$

$L=\frac{0.017}{10.0} \\L=0.0017H$

Now we can apply:

φ=$\frac{L*I}{N}$

φ=$\frac{0.0017*4.30}{483}$

φ=$0.0000151Tm^{2}$